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\title{Chapter 17: Kashiwara's Theorem}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

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% Section 0
%\section{INTRO.}
\begin{frame}{intro. }
    
In this chapter we deal with the direct image under an embedding. 

This will lead us to an important structure theorem for a whole class of $A_n$-modules. 

The noncommutativity of the Weyl algebra will be an essential ingredient in these results. 

In fact, the theorem fails to hold for polynomial rings, as we show at the end of §2. 

Throughout this chapter, the notation of 14.1.2 will be in force.

\end{frame}

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% Section 1
\section{Embeddings}
\begin{frame}[allowframebreaks]{A. }

Let $\iota: X \to X \times Y$ be the standard embedding: $\iota(X) = (X,0)$, where $0$ is the origin of $Y$. 

It follows from Ch. 15, §1, that
\begin{equation}
D_{X \to X \times Y} = \iota^*(A_{m+n})= K[X]\otimes_{K[X,Y]}A_{m+n} \cong A_{m+n}/(Y)A_{m+n},
\end{equation}
where $(Y)$ is the ideal of $K[X,Y]$ generated by the $y$'s. 

One easily verifies that this isomorphism preserves the right $A_{m+n}$-module structure of $\iota^*(A_{m+n})$. 

Transposing this bimodule with the help of the standard transposition of $A_{m+n}$ and using Proposition 16.2.1,
\begin{equation}
D_{X \times Y \leftarrow X} \cong A_{m+n}/A_{m+n}(Y).
\end{equation}

This is an $A_{m+n}-A_n$-bimodule. 

If $N$ is a left $A_n$-module, then its direct image under $\iota$ is, by definition, the module
\begin{equation}
\iota_* N = D_{X \times Y \leftarrow X} \otimes_{A_n} N.
\end{equation}

Let us consider the structure of $D_{X \times Y \leftarrow X}$ in greater detail. 

By Corollary 13.2.3 there exists an isomorphism of bimodules,
\begin{equation}
D_{X \times Y \leftarrow X} \cong (A_m/A_m(Y)) \widehat{\otimes} A_n.
\end{equation}

But $A_m/A_m(Y)$ is isomorphic to $K[\partial_{y_1}, \ldots, \partial_{y_m}] = K[\partial_y]$ as an $A_m$-module. 

Therefore,
\begin{equation}
D_{X \times Y \leftarrow X} \cong K[\partial_y] \widehat{\otimes} A_n.
\label{tag-1.1}
\end{equation}

Let $\alpha \in \mathbb{N}^n$. 

Recall that the action of $y_j$ on $\partial^\alpha \in K[\partial_y]$ is given by
\begin{equation}
y_j \cdot \partial^\alpha = -\alpha_j \partial^{\alpha - e_j}.
\end{equation}

It is worth stressing that on the right-hand side of (\ref{tag-1.1}), the $y$'s and $\partial_y$'s act only on $K[\partial_y]$, whilst the $x$'s and $\partial_x$'s act only on $A_n$; as in Ch. 13, §2. 

Thus if $N$ is a left $A_n$-module,
\begin{equation}
\iota_* N \cong K[\partial_y] \widehat{\otimes} N.
\end{equation}

It is very easy to calculate with the direct image in this form; we give two examples. 

The next theorem should be compared with Theorem 14.3.1.

\textbf{Theorem 17.1.2.}

Let $\iota: X \to X \times Y$ be the standard embedding. 

If $M$ is a finitely generated left $A_n$-module, then:
\begin{enumerate}
    \item $\iota_* M$ is a finitely generated left $A_{m+n}$-module.
    \item $d(\iota_* M) = m + d(M)$.
    \item $m(\iota_* M) \leq m(M)$.
\end{enumerate}

\textbf{Proof:}

(1) follows from Proposition 13.2.1. 

Since $K[\partial_y]$ is the Fourier transform of $K[X]$, it must have dimension $m$ and multiplicity $1$ by Proposition 9.2.2. 

Thus (2) and (3) follow from Theorem 13.4.1.

The following corollary is an immediate consequence of the theorem.

\textbf{Corollary.}

Let $M$ be a {\color{red}holonomic} $A_n$-module and $\iota$ the standard embedding. 

Then $\iota_* M$ is a {\color{red}holonomic} $A_{m+n}$-module.

We now turn to another application.

\textbf{Lemma.}

Let $N$ be a left $A_n$-module. 

Every element of $\iota_* N$ is annihilated by a power of $y_j$, where $1 \leq j \leq m$.

\textbf{Proof:}

Without loss of generality assume that $j=1$. 

Since
\begin{equation}
y_1 \cdot \partial^\alpha = -\alpha_1 \partial^{\alpha - e_1}
\end{equation}
in $K[\partial_y]$, it follows that $\partial^\alpha$ is annihilated by $y_1^k$, for any $k \geq \alpha_1 + 1$. 

Now every element of $\iota_* N$ may be written in the form
\begin{equation}
\sum_{\alpha \in I} (\partial^\alpha \otimes u_\alpha),
\end{equation}
where $u_\alpha \in N$ and $I \subseteq \mathbb{N}^n$. 

Choose an integer $k$ such that $k \geq \alpha_1 + 1$, for every $\alpha \in I$. 

Then $y_1^k \cdot \partial^\alpha = 0$, for every $\alpha \in I$. 

Thus
\begin{equation}
y_1^k \cdot \left(\sum_{\alpha \in I} \partial^\alpha \otimes u_\alpha\right) = 0.
\end{equation}

We also deduce from the isomorphism
\begin{equation}
D_{X \times Y \leftarrow X} \cong K[\partial_y] \widehat{\otimes} A_n
\end{equation}
that $D_{X \times Y \leftarrow X}$ is free of infinite rank as a right $A_n$-module. 

This follows from the fact that $K[\partial_y]$ is a $K$-vector space of infinite dimension. 

The monomials in the $\partial_y$'s form a basis for $D_{X \times Y \leftarrow X}$ as a right $A_n$-module. 

A simple consequence of this fact and Theorem 12.4.2 is the following corollary.

\textbf{Corollary.}

If $N$ is a left $A_n$-module, then $\iota_* N = 0$ if and only if $N = 0$.

\end{frame}

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% Section 2
\section{Kashiwara's Theorem}
\begin{frame}[allowframebreaks]{B. }

In this section we study the direct image under the following special case of the standard embedding $\iota: X \to X \times K$. 

Let $y$ denote the coordinate of $K$. 

The polynomial ring in the $x$'s and $y$ will be denoted by $K[X,y]$. 

Let $M$ be a $K[X,y]$-module and $H$ the hyperplane of equation $y=0$ in $X \times K$. 

The submodule of $M$ of elements with {\color{red}support} on $H$ is defined by
\begin{equation}
\Gamma_H M = \{u \in M : u \text{ is annihilated by a power of } y\}.
\end{equation}

Note that by Proposition 12.5.1, this is exactly the kernel of the map $M \to M[y^{-1}]$. 

Although we have defined $\Gamma_H M$ for $K[X,y]$-modules, in most of our applications it will also be an $A_{n+1}$-module.

\textbf{Proposition.}

Let $M$ be an $A_{n+1}$-module. 

Then $\Gamma_H M$ is a submodule of $M$.

\textbf{Proof:}

Let $u_1, u_2 \in \Gamma_H M$. 

Suppose that $y^{k_1}$ and $y^{k_2}$ are the powers that annihilate $u_1, u_2$, respectively. 

Then $y^{k_1+k_2}$ annihilates both $u_1$ and $u_2$, hence their sum.

It is harder to show that $\Gamma_H M$ is closed under multiplication by elements of $A_{n+1}$. 

Of course it is enough to check that it is closed for multiplication by the generators of $A_{n+1}$. 

Assume that $u \in \Gamma_H M$ is annihilated by $y^k$. 

Note that $y^k$ commutes with the $x$'s, $\partial_x$'s and with $y$. 

Thus,
\begin{equation}
y^k(\partial_{x_i} u) = \partial_{x_i}(y^k u) = 0.
\end{equation}
and so $\partial_{x_i} u \in \Gamma_H M$. 

Similarly, $x_i u, y u \in \Gamma_H M$. 

Hence we need only check that $\partial_y u \in \Gamma_H M$. 

Using the Weyl algebra relations,
\begin{equation}
y^{k+1}(\partial_y u) = \partial_y y^{k+1} u - (k+1) y^k u = 0,
\end{equation}
which completes the proof of the proposition.

Given an $A_{n+1}$-module $M$, put
\begin{equation}
\ker_M(y) = \{u \in M : yu = 0\}.
\end{equation}

Clearly $\ker_M(y) \subseteq \Gamma_H M$. 

Although $\ker_M(y)$ is closed under addition, it is not an $A_{n+1}$-submodule of $M$. 

But, since the $y$'s commute with the $x$'s and $\partial_x$'s, it is an $A_n$-submodule of $M$. 

We now make explicit the true relation between $\ker_M(y)$ and $\Gamma_H M$. 

It all begins with a lemma. 

To simplify the notation, let $M_0 = \ker_M(y)$.

\textbf{Lemma.}

Let $M$ be a left $A_{n+1}$-module, and let $H$ be the hyperplane $y=0$. 

Then

\begin{enumerate}
    
  \item For each $k \in \mathbb{N}$, the map
    \begin{equation}
    \partial_y: \partial_y^k M_0 \to \partial_y^{k+1} M_0
    \end{equation}
    defined by right multiplication by $\partial_y$ is injective.
    
  \item $y(A_{n+1} M_0) = A_{n+1} M_0$.
    
  \item $A_{n+1} M_0 = M_0 \oplus \partial_y M_0 \oplus \partial_y^2 M_0 \oplus \ldots$
    
\end{enumerate}

\textbf{Proof:}

Let $u \in M_0$ and $k$ be a positive integer. 

We have that $[y, \partial_y^k] = -k \partial_y^{k-1}$ and that $y$ annihilates $u$, thus $y \cdot \partial_y^k u = -k \partial_y^{k-1} u$. 

Therefore,
\begin{equation}
y^i \cdot \partial_y^k u = (-1)^i k(k-1)\cdots(k-i+1) \partial_y^{k-i} u
\tag{2.3}
\end{equation}
for every $i \geq 0$. 

Suppose that $\partial_y u = 0$. 

Applying (2.3) with $i=k$ to this equation, we conclude that $u=0$. 

This proves (1).

Now, every element of $A_{n+1}$ can be written in the form
\begin{equation}
\sum_{0}^{k} \partial_y^i b_i,
\end{equation}
where $b_i \in A_n[y]$. 

Since $M_0$ is a left $A_n$-module that is annihilated by $y$, the elements of $A_{n+1} M_0$ can be written as
\begin{equation}
\sum_{0}^{k} \partial_y^i u_i
\end{equation}
where $u_i \in M_0$. 

In other words,
\begin{equation}
A_{n+1} M_0 = M_0 + \partial_y M_0 + \partial_y^2 M_0 + \ldots
\end{equation}

Thus (2) follows from $\partial_y^k u = -(k+1)^{-1} y \cdot (\partial_y^{k+1} u)$.

Assume now that
\begin{equation}
u_0 + \partial_y u_1 + \cdots + \partial_y^k u_k = 0,
\end{equation}
where $u_0, \ldots, u_k \in M_0$. 

Multiplying by $y^k$ and using (2.3) we conclude that $u_k = 0$. 

Thus we have a sum with $k-1$ terms. 

By induction on the number of terms, we conclude that $u_0 = \cdots = u_k = 0$. 

Hence the sum is direct, and (3) is proved.

We are now ready to state a preliminary version of Kashiwara's theorem.

\textbf{Theorem.}

Let $M$ be a left $A_{n+1}$-module, let $H$ be the hyperplane $y=0$ and denote by $\iota: X \to X \times K$ the standard embedding. 

The $A_{n+1}$-modules $\iota_*(\ker_M y)$ and $\Gamma_H M$ are isomorphic.

\textbf{Proof:}

We have seen that $M_0 = \ker_M y$ is an $A_n$-submodule of $M$. 

It follows from §1 that
\begin{equation}
\iota_*(M_0) = K[\partial_y] \widehat{\otimes} M_0.
\end{equation}

We wish to show that this is isomorphic to $\Gamma_H M$.

Since $M_0 \subseteq \Gamma_H M$, we can use the universal property of the tensor product to define a map
\begin{equation}
\phi: K[\partial_y] \widehat{\otimes} M_0 \to \Gamma_H M
\end{equation}

by $\phi(f \otimes u) = f u$, where $f$ is a polynomial in $\partial_y$. 

The image of $\phi$ is $A_{n+1} M_0$ and its kernel is zero, by Lemma 2.2(3). 

Thus to prove that $\phi$ is surjective it is enough to show that $\Gamma_H M$ is contained in the $A_{n+1}$-submodule of $M$ generated by $M_0$.

Let $u \in \Gamma_H M$ and assume that $y^k u = 0$. 

We want to show that $u \in A_{n+1} M_0$. 

The proof is by induction on $k$. If $k=1$, then $u \in M_0$. 

Suppose that the result holds for every element of $M_0$ annihilated by $y^{k-1}$. 

From $y^k u = 0$, we deduce that $\partial_y(y^k u) = 0$. Since $[\partial_y, y^k] = k y^{k-1}$,

\begin{equation}
y^{k-1}(k u + y \partial_y u) = 0.
\end{equation}

Thus, by the induction hypothesis, $k u + y \partial_y u \in A_{n+1} M_0$. But $y^{k-1}(y u) = y^k u = 0$ and so $y u \in A_{n+1} M_0$. Since $A_{n+1} M_0$ is an $A_{n+1}$-submodule of $M$, we have that

\begin{equation}
k u + y \partial_y u - \partial_y y u \in A_{n+1} M_0.
\end{equation}

But this expression is equal to $(k-1) u$, because $[\partial_y, y] = 1$. 

Thus $u \in A_{n+1} M_0$, as required.

Combining the fact that $\Gamma_H(M) = A_{n+1} M_0$ with Lemma 2.2(2) we get the following result, which will be used in the next chapter.

\textbf{Corollary.}

Let $M$ be a left $A_{n+1}$-module and let $H$ be the hyperplane $y=0$. 

Then $y \Gamma_H(M) = \Gamma_H(M)$.

We say that a $K[X,y]$-module $M$ has support on $H$ if $\Gamma_H M = M$. 

In particular, this definition may be applied to $A_{n+1}$-modules. 

The next result is a structure theorem for $A_{n+1}$-modules with support on $H$. 

The proof is an immediate consequence of Theorem 2.4 and the above definition.

\textbf{Corollary.}

Let $M$ be an $A_{n+1}$-module with support on $H$. Then $M \cong \iota_*(\ker_M y)$.

These results hold in greater generality. 

In Ch. 18, §3 we will use the language of category theory to bring forth the true nature of Kashiwara's theorem.

We have taken care to point out that both $\Gamma_H M$ and $\ker_M y$ are defined for $K[X,y]$-modules as well as for $A_{n+1}$-modules. 

In fact there is also a natural way to define direct images for modules over polynomial rings under polynomial maps; see Exercise 3.4. 

Applied to the standard embedding $\iota: X \to X \times K$, the direct image $\iota_* M$ of a $K[X]$-module $M$ equals $M$, as a vector space. 

The action of $f(X,Y) \in K[X,y]$ on $u \in M$ is given by $f u = f(X,0) u$.

For this direct image, it is not true that if $M$ is a $K[X,y]$-module then

\begin{equation}
\iota_*(\ker_M y) \cong \Gamma_H M.
\end{equation}

Here is a simple example. 

Consider the $K[X,y]$-module $M = K[X,y]/(y^2)$. 

Since every element of $M$ is annihilated by $y^2$, it follows that $\Gamma_H M = M$. 

But a straightforward calculation shows that $\ker_M y = y M$. 

Since $\iota_*(\ker_M y) = y M$, as vector spaces, we have that $\iota_*(\ker_M y) \neq \Gamma_H M$.

\end{frame}

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% Section 3
\section{Exercises}
\begin{frame}[allowframebreaks]{C. }

\textbf{Exercise 3.1.}

Let $\iota: K \to K^2$ be the standard embedding given by $\iota(x) = (x,0)$. 

Compute the direct image of the following modules under $\iota$.
\begin{enumerate}
    \item $K[x]$
    \item $A_1/A_1 \delta^5$
    \item $A_1/A_1 x \partial$
    \item $A_1/A_1 (x^2 \partial + 4x + 1)$
\end{enumerate}

\newpage

\textbf{Exercise 3.2.}

Let $\iota: X \to X \times Y$ be the standard embedding. 

Let $M$ be a left $A_n$-module. 

Is it true that
\begin{equation}
\iota_*(M_\mathcal{F}) \cong (\iota_* M)_\mathcal{F}?
\end{equation}


\newpage

\textbf{Exercise 3.3.}

Let $\iota: X \to X \times K$ be the standard embedding and let $M$ be a left $A_{n+1}$-module with support on the hyperplane $H$ of equation $y=0$. 

Show that $\ker_M y$ is isomorphic, as a left $A_n$-module, to
\begin{equation}
\operatorname{Hom}_{A_{n+1}}(D_{X \times K \leftarrow X}, M).
\end{equation}


\newpage

\textbf{Exercise 3.4.}

The purpose of this exercise is to define a direct image for modules over polynomial rings. 

Suppose that $F: X \to Y$ is a polynomial map. 

The comorphism of $F$ is a homomorphism of rings $F^\sharp: K[Y] \to K[X]$. 

Let $M$ be a $K[X]$-module. 

We want to construct a $K[Y]$-module $F_+ M$. 

The definition is as follows. 

As $K$-vector spaces, $F_+ M = M$. 

The action of $g \in K[Y]$ on $u \in M$ is defined by $g u = F^\sharp(g) u$.

\begin{enumerate}
    \item Check that this action makes $F_+ M$ into a $K[Y]$-module.
    \item Compute $\iota_+ M$ for the standard embedding $\iota: X \to X \times Y$.
    \item Let $N$ be a $K[X,Y]$-module. Compute $\pi_+ N$ for the projection $\pi: X \times Y \to Y$ onto the second coordinate.
    \item If $M$ is finitely generated over $K[X]$, is $F_+ M$ necessarily finitely generated over $K[Y]$?
\end{enumerate}


\newpage

\textbf{Exercise 3.5.}

Let $M$ be a left $A_n$-module and $\phi: M \to M[x_n^{-1}]$ the natural embedding. 

Show that $\ker \phi$ and $\operatorname{coker} \phi$ are supported on the hyperplane $x_n=0$.

\newpage

\textbf{Exercise 3.6.}

Let $M$ be an irreducible $A_n$-module. 

For a non-zero element $u$ in $M$ let $J(u) = \{f \in K[X]: f u = 0\}$. Show that:
\begin{enumerate}
    \item If $0 \neq v \in M$ then $\operatorname{rad} J(v) = \operatorname{rad} J(u)$.
    \item $\operatorname{rad} J(u)$ is a prime ideal of $K[X]$.
    \item If $M$ is supported on the hypersurface $x_n=0$, then $\operatorname{rad} J(u)$ is contained in the ideal of $K[X]$ generated by $x_n$.
\end{enumerate}




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